Reputation: 7838
I have a bunch of Jpg images in byte array form. I want to add these to a zip file, turn the zip file into a byte array, and pass it to somewhere else. In a method, I have this code:
var response = //some response object that will hold a byte array
using (var ms = new MemoryStream())
{
using (var zipArchive = new ZipArchive(ms, ZipArchiveMode.Create, true))
{
var i = 1;
foreach (var image in images) // some collection that holds byte arrays.
{
var entry = zipArchive.CreateEntry(i + ".jpg");
using (var entryStream = entry.Open())
using (var compressStream = new MemoryStream(photo.ImageOriginal))
{
compressStream.CopyTo(entryStream);
}
i++;
}
response.ZipFile = ms.ToArray();
}
using (var fs = new FileStream(@"C:\Users\MyName\Desktop\image.zip", FileMode.Create))
{
ms.Position = 0;
ms.CopyTo(fs);
}
}
return response;
Now, I've added a filestream near the bottom to write it to a zipfile right away for testing purposes. This works, I get a zip file with 1 or more images in it on my desktop. However: response.ZipFile
can not be made into a valid zipfile in the same way. I have tried this:
using (var ms2 = new MemoryStream(response.ZipFile))
using (var fs = new FileStream(@"C:\Users\Bara\Desktop\image.zip", FileMode.Create))
{
ms2.Position = 0;
ms2.CopyTo(fs);
}
But that creates a zipfile that can not be opened.
What I'm trying to do: turn response.ZipFile
into an array that can be turned into a working zipfile again. What am I doing wrong in this code?
Upvotes: 0
Views: 1231
Reputation: 44201
How do you know that ZipArchive
's Dispose
doesn't write more to the underlying stream?
You should move this line to be after disposing the ZipArchive
:
response.ZipFile = ms.ToArray();
Full code:
var response = //some response object that will hold a byte array
using (var ms = new MemoryStream())
{
using (var zipArchive = new ZipArchive(ms, ZipArchiveMode.Create, true))
{
var i = 1;
foreach (var image in images) // some collection that holds byte arrays.
{
var entry = zipArchive.CreateEntry(i + ".jpg");
using (var entryStream = entry.Open())
using (var compressStream = new MemoryStream(photo.ImageOriginal))
{
compressStream.CopyTo(entryStream);
}
i++;
}
}
response.ZipFile = ms.ToArray();
}
return response;
Upvotes: 2